August 21, 2011

Problem 302 Van Khea

Filed under: Problem by Van Khea — KKKVVV @ 3:34 pm

If a, b, c are positive real numbers then prove that:

\displaystyle \frac{a^2}{b\sqrt{c^2+3ab}}+\frac{b^2}{c\sqrt{a^2+3bc}}+\frac{c^2}{a\sqrt{b^2+3ca}}\geq \frac{3}{2}
We have
\displaystyle \frac{a^2}{b\sqrt{c^2+3ab}}+\frac{b^2}{c\sqrt{a^2+3bc}}+\frac{c^2}{a\sqrt{b^2+3ca}}\displaystyle =\frac{(a^2)^{\frac{3}{2}}}{\sqrt{a^2b^2c^2+3(ab)^3}}+\frac{(b^2)^{\frac{3}{2}}}{\sqrt{a^2b^2c^2+3(bc)^3}}+\frac{(c^2)^{\frac{3}{2}}}{\sqrt{a^2b^2c^2+3(ca)^3}}
Because \displaystyle \frac{3}{2}-\frac{1}{2}=1 then we have
\displaystyle \frac{(a^2)^{\frac{3}{2}}}{\sqrt{a^2b^2c^2+3(ab)^3}}+\frac{(b^2)^{\frac{3}{2}}}{\sqrt{a^2b^2c^2+3(bc)^3}}+\frac{(c^2)^{\frac{3}{2}}}{\sqrt{a^2b^2c^2+3(ca)^3}}\displaystyle \geq \frac{(a^2+b^2+c^2)^{\frac{3}{2}}}{\sqrt{3((ab)^3+(bc)^3+(ca)^3+a^2b^2c^2)}}
Thus, it suffices to show that
\displaystyle \frac{(a^2+b^2+c^2)^{\frac{3}{2}}}{\sqrt{3((ab)^3+(bc)^3+(ca)^3+a^2b^2c^2)}}\geq \frac{3}{2}
\displaystyle \Leftrightarrow 27((ab)^3+(bc)^3+(ca)^3+a^2b^2c^2)\leq 4(a^2+b^2+c^2)^3
Letting a^2=x; b^2=y; c^2=z yields
\displaystyle 27((xy)^{\frac{3}{2}}+(yz)^{\frac{3}{2}}+(zx)^{\frac{3}{2}}+xyz)\leq 4(x+y+z)^3
From Cauchy-Schwarz inequality we have
\displaystyle ((xy)^{\frac{3}{2}}+(yz)^{\frac{3}{2}}+(zx)^{\frac{3}{2}}+xyz)^2\leq (x^2y+y^2z+z^2x+xyz)(xy^2+yz^2+zx^2+xyz)
So we need to prove that
\displaystyle (x^2y+y^2z+z^2x+xyz)(xy^2+yz^2+zx^2+xyz)\leq \frac{16}{27^2}(x+y+z)^6
Now we will show that for x, y, z are positive real numbers then we have
27(x^2y+y^2z+z^2x+xyz)\leq 4(x+y+z)^3 ;(1) and 27(xy^2+yz^2+zx^2+xyz)\leq 4(x+y+z)^3 ; (2)
without loss of generality, suppose that x=min(x, y, z). Sitting y=x+u and z=x+v; (u, v\geq 0) then:
(1)\Leftrightarrow 9(u^2-uv+v^2)x+(u-2v)^2(4u+v)\geq 0
(2)\Leftrightarrow 9(u^2-uv+v^2)x+(2u-v)^2(u+4v)\geq 0
which is obviously true. Equality occurs for u=v=0\Leftrightarrow x=y=z
\displaystyle \Rightarrow ((xy)^{\frac{3}{2}}+(yz)^{\frac{3}{2}}+(zx)^{\frac{3}{2}}+xyz)^2\displaystyle \leq \frac{16}{27^2}(x+y+z)^6
\displaystyle \Leftrightarrow 27((xy)^{\frac{3}{2}}+(yz)^{\frac{3}{2}}+(zx)^{\frac{3}{2}}+xyz)\leq 4(x+y+z)^3 is true.
Therefore the proof is completed. Equality occurs for x=y=z\Leftrightarrow a=b=c


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